Chapter 5 Turning Effect of Forces

Chapter 5 Thrning Effect of Forces

Learning Outcomes

After completing this chapter, students should be able to:

1. describe the moment of a force in terms of its turning effect and relate this to everyday example 2. recall and apply the relationship moment of a force (or torque) = force x perpendicular distance from

the pivot to new situations or to solve related problems 3. state the principle of moments for a body in equilibrium 4. apply the principle of moments to new situations or to solve related problems 5. show understanding that the weight of a body may be taken as acting at a single point known as its

centre of gravity 6. describe qualitatively the effect of the position of the centre of gravity on the stability of the objects

5.1 Moments page 87 1. Lesson may be introduced by asking students to list down situations involving turning forces.

Examples include opening a tap, tightening a nut, opening a door, turning a car steering, opening tins with screwdrivers and spoons etc.

2. Ask students why the handle on a door is placed a long distance away from the pivot.

3. Emphasise to students the relationship moment offorce (or torque) = force x perpendicular distance from the pivot. The perpendicular distance is measured from the line of action of the force to the pivot.

4. Discuss qualitatively and give examples so that students understand what is meant by clockwise moments and what is meant by anticlockwise moments.

5. Students are likely to be familiar with children of unequal weights balancing on see-saws. Ask how it is done.

6. Explain the meaning or equilibrium, i.e. (a) the object does not move linearly and (b) the object does not rotate (no resultant moments).

7. Help students to recall what they have studied in primary school. Levers are not mentioned in the syllabus but they are examples or the application of the moment of a force.

Answer to Think Time question . Page 88

Use the screw driver that has a fat handle. The torque will be greater.

Answers to Section Review questions page 91

1. Moment = Force x perpendicular distance = 300 N x 0.15 m = 45 N m

It is easier to undo the nut with a longer spanner because the moment produce will be greater.

2. Wx 1.5 = (45 x 1.0) + (20 x 1.5) W = 50N

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5.2 Centre of Gravity 1. You may start the lesson by asking the following questions:

(a) Why does the Leaning Tower of Pisa not topple over?

Chapter 5 Turning Effect of Forces

page 92

(b) Why is it impossible to stand with your back and your heels against a wall and bend over to touch your toes without toppling forward?

(a) The centre of gravity of the Leaning Tower of Pisa lies above the base of support, so the tower will not topple.

Centre of gravity

Base of support

(b) You can bend over to touch your toes without toppling forward only if your centre of gravity is above the area bounded by your feet.

Centre of gravity

)l

2. Emphasise that the centre of gravity of a body is a single point through which the entire weight of an object can be taken to act. If a body is hanging freely at rest, then its centre of gravity is always vertically below the pivot.

3. Ask the student to hold two rods of different lengths vertically and allow both to fall at the same time. Which rod should hit the ground flrst? Explain.

Centre of gravity

The shorter rod always hits the ground first. The CO of the longer rod is vertically higher above the ground than that of the shorter rod, which means that the longer rod will take longer time to fall to the ground.

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Chapter 5 Turning Effect of Forces

4. Securely fasten a hammer to a metre rule with the piece of string. Adjust the position of the knot on the hammer until the handle of the hammer touches the ruler. The contraption may now be balanced as shown below. Explain how the balancing can occur.

~' TableTop

The hammerhead is the heavier part of the hammer. With the arrangement shown, the e.G. of the system is below the support on the tabletop, thus balancing is possible.

5. In the experiment to determine the centre of gravity on page 94, students must ensure that the friction between the pin and the holes of the lamina is kept to a minimum.

Answer to Think Time question Page 93

If the centre of gravity is not directly below the pivot, there will be a moment turning the object.

5.3 Stability page 93 1. The stability of a body is measured by the angle through which it can be tilted before toppling over.

The greater the angle needed, the more stable it is.

2. A body will topple over when the vertical line through its centre of gravity falls outside its base.

3. It is possible to make a body more stable by (a) lowering its centre of gravity, (b) increasing the area of its base.

4. Ask students to give examples of making a body more stable.

5. Discuss the effect of heavy chassis on the stability of buses or the results of heavy loads on the roofs of narrow minibuses.

Answer to Think Time question Page 94

The filing cabinet is in danger of toppling over because the CG may extend beyond the support base.

Physics in Daily Life: Balancing Moments and Restoring Torques page 96 Answers to Q

1. A marble released on the edge of a bowl will oscillate with decreasing amplitude until it settles down at the bottom of the bowl.

2. (a) The period of a pendulum is dependent on its length and gravitational acceleration g. (b) The period decreases quickly if damping increases. (c) Assuming the length of a pendulum and g remains constant, the period will not change if

damping is absent.

Answers to Section Review questions page 97

1. Her centre of gravity is somewhere above the foot in contact with the beam.

2. This increases their stability. When the upper part of the body moves, the vertical line through the CG will always be between the two feet, so they will not be easily toppled over by an opponent.

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Chapter 5 Turning Effect of Forces

3. Doughnut - e.G. is in the centre of the ring Basketball- e.G. is at the centre of the ball Waste paper basket - e.G. is inside the pot, it is somewhere near the centre Boomerang - C.G. is at the elbow of the boomerang

4. Truck A. This is because the vertical line through the centre of gravity falls out of the base (outside the wheel) .

Answers to Misconception Analysis page 98,99

1. False. 2. True 3. True 4. True 5. False. 6. True 7. True 8. False. 9. False. 10. True

Moment = force x perpendicular distance from the pivot

Centre of gravity may be outside the body.

The object can still topple but it topples after turning a greater angle. It depends on where the CG lies.

Answers to Multiple Choice Questions page 99,100

1. B 2. A 3. C 4. 5.

B A

clockwise moment = 4 N x 1.5 m = 6.0 Nm; anticlockwise moment = 5 N x 0.8 m = 4.0 m (0.4 N x 70 cm) + (W x 20 cm) = 1.0 N x 30 cm

6. C

7. B

8. B

9. C 10. B

W = 0.1 N (F xl m) + (12 x 2 m) = 20 N x 4 m

F = 16N F x 4 m = 60 N x 2 m

F = 30N F x 40 cm = 200 N x 5 cm

F = 25 N

Answers to Structured Questions page 101, 102

1. 5 N x 20 cm = Wx 50 cm W=2N

2. The weight of the metre rule will act through the 50 cm mark Wx30cm=2.4 x15cm

W = 1.2 N

3. F x 2 m = 600 N x l.5 m F= 450 N

4. 200 N x x = 50 x 20 cm x = 5 cm from the centre

5. Let the support be I cm from the 60-N weight. So the support will be (20 -I) cm from the 40 N weight 60 N x I cm = 40 N x (20 -I) em

1= 8 cm from the 60 N weight

6. The moment caused by the unknown weight and the moment caused by the l.2-N weight are the

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same. W x 30 cm = 1.2 N x 45 cm

W = 1.8 N

7. Fx3 .0m = 600x4.0m F = 800N

8. (a) 1.5Nx40cm = (2.0Nx20cm)+(0.5Nxx) x = 40 cm from the pivot on the side of 2.0 N weight

(b) Move the pivot towards the 1.5 N weight

9. T x 100 cm = 30 N x 60 cm T = 18N

Chapter 5 Turning Effect of Forces

If the string is attached to any other point on the trapdoor between A and B, the force required will be greater because the distance from the pivot is smaller.

10. The weight 3.0 N of the metre rule will act through the 50 cm mark (Wx 60 cm) + (3.0 N x 10 cm) = 7.0 N x 30 cm

W = 3.0N

11. The weight 10 N of the uniform rod will act through the 50 cm mark W x 30 cm = (10 N x 20 cm) + (8 N x 35 cm)

W = 16N

12. (a) The weight 10 N of the uniform rod will act through the 50 cm mark (10 N x 50 cm) + (Wx 70 cm) = 19 N x 100 cm

W 20 (b) m = - = - = 2 kg

g 10

W = 20N

Answers to Critical Thinking Questions . page 101, 102

1. The torque remains the same even if a length of rope is fastened to the wrench. An alternative solution is to lengthen the arm of the wrench by placing a water pipe over it.

2. The forces are not necessary the same. Moment also depends on the perpendicular distance from the pivot to the force.

3. You should rest the CG of the ladder on the shoulder. In this way there will not be any torque trying to turn the ladder.

4. The centre of gravity is vertically above the base of the rack.

5. This is because the centre of gravity changes position as the tub spins.

6. It is possible to balance according to the arrangement shown because the centre of gravity is vertically below the point of support. Any disturbance of the system will cause the torque to tum the arrangement back to its balance position.

Extension page 102

In Figure 5.45, once the ship is tipped to the left, the weight and the upthrust will produce anticlockwise moment which will cause further tipping. The ship is not stable in this situation.

So for stability, the CG of the ship has to be as low as possible. Heavy equipment, such as the ship's engines, should therefore be mounted low in the hull, and massive cargo should be stored in holds near the keeL

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